MHB How Do You Find the Multiplicative Inverse in \( Z_{360}[x] \)?

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To find the multiplicative inverse of the polynomial \( f(x) = 30x^4 + 120x^2 + 240x + 7 \) in \( Z_{360}[x] \), one approach is to set \( f^{-1}(x) = a_n x^n + ... + a_0 \) and establish congruences. The key equation is \( 7a_0 \equiv 1 \pmod{360} \), which can be solved using the Euclidean algorithm or by inspection. Through inspection, it is determined that \( 7^{-1} \equiv 103 \pmod{360} \). This method indicates that finding the inverse can be simplified with a few calculations rather than relying solely on the Euclidean algorithm.
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How would you find the multiplicative inverse of the following?

Let $R=Z_{360}[x]$
Find the multiplicative inverse of $f(x)=30x^4+120x^2+240x+7$ in $R$.

Do you have to solve it using the Euclidean Algorithm? If so, I'm not sure how to do that.
This problem has me stumped.

Any help is much appreciated.
Thanks!
 
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corkycorey101 said:
How would you find the multiplicative inverse of the following?

Let $R=Z_{360}[x]$
Find the multiplicative inverse of $f(x)=30x^4+120x^2+240x+7$ in $R$.

Do you have to solve it using the Euclidean Algorithm? If so, I'm not sure how to do that.
This problem has me stumped.

Any help is much appreciated.
Thanks!

Hi corkycorey101! Welcome to MHB! ;)

Let the multiplicative inverse be $f^{-1}(x)=a_n x^n + ... + a_0$.
Then:
$$7a_0 \equiv 1 \pmod{360}\\
240a_0 + 7a_1 \equiv 0 \pmod{360}\\
...
$$

To solve $7a_0 \equiv 1$, we could apply the Euclidean algorithm, but let's try inspection first.
If we divide $360 + 1$ by $7$, the remainder is $4$.
If we divide $2\cdot 360 + 1$ by $7$, the remainder is $0$.
So:
$$7^{-1} \equiv \frac{2\cdot 360 + 1}{7} \equiv 103 \pmod{360}$$
Generally, small as $7$ is, we can be sure to find sufficient information within 3 tries.
 
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